document.write( "Question 1133805: Find the maximum integral value of k such that 2014^k divides (1! x 2! x 3!x ... x 2014!)? \n" ); document.write( "

Algebra.Com's Answer #751124 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "The prime factorization of 2014 is 2*19*53. So

\n" ); document.write( " $\"2014%5Ek+=+%282%5Ek%29%2819%5Ek%29%2853%5Ek%29\"$

\n" ); document.write( "We are to find the maximum value of k for which 2014^k divides (1!*2!*3!*...*2014!).

\n" ); document.write( "In that product of consecutive factorials, the smallest prime factor of 2014 will occur the most often, and the largest prime factor will occur least often. So the largest value of k for which 2014^k divides that product of factorials is the number of factors of 53 in that product of factorials.

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document.write( "The 1st set of 52 factorials -- 1! through 52! -- contain no factors of 53.\r\n" );
document.write( "The 2nd set of 53 factorials -- 53! through 105! -- each contain 1 factor of 53  -->  53 factors of 53.\r\n" );
document.write( "The 3rd set of 53 factorials -- 106! through 158! -- each contain 2 factors of 53  -->  53*2 = 106 factors of 53.\r\n" );
document.write( "The 4th set of 53 factorials -- 159! through 211! -- each contain 3 factors of 53  -->  53*3 = 159 factors of 53.\r\n" );
document.write( "...\r\n" );
document.write( "...\r\n" );
document.write( "The 37th set of 53 factorials -- 1961! through 2013! -- each contain 37 factors of 53  -->  53*37 = 1961 factors of 53.\r\n" );
document.write( "The last factorial -- 2014! -- contains 38 factors of 53.

\n" ); document.write( "The total number of factors of 53 in the product of the consecutive factorials is

\n" ); document.write( "53(1+2+3+...+37)+38 = 37297

\n" ); document.write( "ANSWER: The product of factorials is divisible by 2014^37297 but not by 2014^37298.
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